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Author Topic: Reactive Generator Research for everyone to share  (Read 64439 times)

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It's not as complicated as it may seem...
I'd prefer if you did post ION. It's lonely out here sometimes :(

 :D
   
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I'd prefer if you did post ION. It's lonely out here sometimes :(

 :D

Hi 99,

Are you planning to build a test bed for like the one Luc did?

GL.
   
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Reactive power does no useful work in the electrical load and does not load the source.
All it takes to believe this are simple experiments and the ability to correctly interpret the results.

This is probably the most repeated 'OH! I've just discovered something NEW!' to be seen in these parts.

'Reversing the scope probes'?

Even if that act could be useful, and IT IS NOT, do these folks know how to invert the signal? (Obviously not).

.99, you are beating your forehead against another rosie. Anyone with a basic understanding of AC electricity would consider their claims of your mistakes as being laughable.
   

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It's not as complicated as it may seem...
Hi 99,

Are you planning to build a test bed for like the one Luc did?

GL.
GL.

As a matter of fact, yes I do. I ordered the 3 parts today.

Why do you ask?
   

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It's not as complicated as it may seem...
'Reversing the scope probes'?

Even if that act could be useful, and IT IS NOT, do these folks know how to invert the signal? (Obviously not).
Reverse the current signal via the scope, yes. Why? because we all place our probes in series-opposition, and as such the computed power comes out the reversed polarity. So to correct the issue, I am simply proposing that in critical situations where a claim hinges on the computed polarity of the power, that the current probe channel be inverted.

Let me repeat the mantra that so many refuse to accept:

Power sources compute to a negative power, and loads compute to a positive power.

This of course assumes that you are not measuring a OU device, and that your measurement is significantly free from parasitic reactances.
   
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GL.

As a matter of fact, yes I do. I ordered the 3 parts today.

Why do you ask?

99,

Just curious. :-)

I will be testing also. Ordered a 200 Watt audio amplifier today.
Will use a toroid transformer on the output of the amplifier to get a "mains".
Will use two 12 Volt batteries on the input of the amplifier.
Much more easy to measure the input going to the system since it will be DC.

GL.
   

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It's not as complicated as it may seem...
The only potential issue there GL is that you are introducing another load (the amp) and it will have its own dissipation to be accounted for. The amp will have a quiescent power, plus a dynamic power dissipation depending on the amount of power it is putting to the load. ION can probably provide some insight to this.

So I think you still end up needing to measure the AC power supplied by the amp at its output. Don't you?
   
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99,

Just curious. :-)

I will be testing also. Ordered a 200 Watt audio amplifier today.
Will use a toroid transformer on the output of the amplifier to get a "mains".
Will use two 12 Volt batteries on the input of the amplifier.
Much more easy to measure the input going to the system since it will be DC.

GL.

I'm sure you realize that the amplifier and it's internal switching supply will dissipate some power, and this will not be a simple task to separate from delivered power unless you characterize your amplifier by running it with different resistive loads at various power levels and plotting total used power,  power dissipated by the amp, and power dissipated by the load. Frequencies above 10kHz will probably also affect the curves, so the characterization needs to be done at perhaps a minimum of three different frequencies.

Losses in your toroidal transformer should probably be tallied also.

Once characterized you will then be able to relate power input, internal losses, and power output, and have a useful test amplifier.

Poynt: I typed this before I saw your post, we are on the same page.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
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The only potential issue there GL is that you are introducing another load (the amp) and it will have its own dissipation to be accounted for. The amp will have a quiescent power, plus a dynamic power dissipation depending on the amount of power it is putting to the load. ION can probably provide some insight to this.

So I think you still end up needing to measure the AC power supplied by the amp at its output. Don't you?

99,

Yes.

There will also be losses in the output transformer connected to the amplifier. So I will need to put in
a shunt resistor as you have shown in your drawings and measure the way you have shown in your simulations.

The main reason for using a amplifier is that I can not mess with the mains where I live now.

GL.
   
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I'm sure you realize that the amplifier and it's internal switching supply will dissipate some power, and this will not be a simple task to separate from delivered power unless you characterize your amplifier by running it with different resistive loads at various power levels and plotting total used power,  power dissipated by the amp, and power dissipated by the load. Frequencies above 10kHz will probably also affect the curves, so the characterization needs to be done at perhaps a minimum of three different frequencies.

Losses in your toroidal transformer should probably be tallied also.

Once characterized you will then be able to relate power input, internal losses, and power output, and have a useful test amplifier.

Poynt: I typed this before I saw your post, we are on the same page.

ION,

Agree to that.

Added: The amplifier does not have a internal switching supply. Circuit diagram attached.
           I will run the amplifier from two 12 volt batteries.

GL.
   

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It's not as complicated as it may seem...
GL.

 O0
   

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It's not as complicated as it may seem...
GL,

As that amp was designed to run on +/- 40V, +/- 12V is almost 1/4 the voltage, and you may not have enough bias adjust left to get the output devices biased up correctly.
   
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Reverse the current signal via the scope, yes. Why? because we all place our probes in series-opposition, and as such the computed power comes out the reversed polarity. So to correct the issue, I am simply proposing that in critical situations where a claim hinges on the computed polarity of the power, that the current probe channel be inverted.

Let me repeat the mantra that so many refuse to accept:

Power sources compute to a negative power, and loads compute to a positive power.

This of course assumes that you are not measuring a OU device, and that your measurement is significantly free from parasitic reactances.

Yes, agreed.

My mention of channel inversion was because it is clear that if Luc thought reversing the scope probe connection would provide some benefit then he should learn how to invert the signal instead of physically reversing the scope probe tip & ground clip connection.

Let us hope for his sake that his scope or probes are of the type that would allow him to physically swap the probe tip and ground clip without shorting something.

Series opposing is the only way to go, when in doubt. If any other arrangement causes problems then the experimenter should start doubting his understanding of his equipment and his circuit.  ;)
   
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GL
With +/- 12 volts supply you will have a lot less than 200 watts. Additionally the amp is class B with some idle current (sometimes called AB) so the output transistors will dissipate power beyond and above the idle current if driven with sine waves.

In other words, the quiescent current is not the only thing to subtract from the input power. The amp must be characterized dynamically.

If you try to get more power by presenting a low impedance e.g. 2 ohm impedance transformer primary, you may exceed the transistor ratings.

Characterization still applies, even in the absence of an internal switcher supply.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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It's not as complicated as it may seem...
He is using series-opposing WW. As most of us do that don't have expensive current probes.

The problem is that he refuses to accept that in doing so, the polarity of the computed power will be backwards, and he is relying heavily on that negative number for his claim that not only is he powering a load, but he is returning excess energy back to the grid.  C.C
   
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GL,

As that amp was designed to run on +/- 40V, +/- 12V is almost 1/4 the voltage, and you may not have enough bias adjust left to get the output devices biased up correctly.

99,

I hope I do. If not, then I use different valued resistors around the bias transistor.
Just a engineering problem. :-)

GL.
   
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GL
With +/- 12 volts supply you will have a lot less than 200 watts. Additionally the amp is class B with some idle current (sometimes called AB) so the output transistors will dissipate power beyond and above the idle current if driven with sine waves.

In other words, the quiescent current is not the only thing to subtract from the input power.

If you try to get more power by presenting a low impedance e.g. 2 ohm impedance transformer primary, you may exceed the transistor ratings.

Characterization still applies, even in the absence of an internal switcher supply.

ION,

Much less than 200 Watt. Yes, that was the plan.

Transistors can handle 20 Amp peak. Maybe not enough for 4 Ohm load and 2x12 Volt DC input?

GL.
   
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The amp is rated for approx 8 ohms (40x40)/8=200 Watts peak power.

At +/- 12 volts it is (12x12)/8= 18 Watts

Of course this is a rough guess, actually there are diode drops, resistive losses, transistor junction drops that reduce this, especially at lower voltages not to mention derating to RMS.

It is more likely that the amp with a +/- 40 volt supply is rated 200 W RMS (not peak) into 4 ohms considering all losses and derating from peak power.

Which is why high power auto amps have boost switcher / inverters built in.



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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
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The amp is rated for approx 8 ohms (40x40)/8=200 Watts peak power.

At +/- 12 volts it is (12x12)/8= 18 Watts

Of course this is a rough guess, actually there are diode drops, resistive losses, transistor junction drops that reduce this, especially at lower voltages not to mention derating to RMS.

It is more likely that the amp with a +/- 40 volt supply is rated 200 W RMS (not peak) into 4 ohms considering all losses and derating from peak power.

Which is why high power auto amps have boost switcher / inverters built in.



ION,

Amplifier Features:

200W music power @ 4 ohm load
100Wrms power @ 4 ohm load
 70Wrms power @ 8 ohm load
Distortion: 0.02% @ 1KHz/10W
Damping factor: >800
Frequency response: 3Hz to 200KHz (-3dB)
Sensitivity: 0.6Vrms
Signal to noise ratio: 115dB
Power supply: 2 x 25-30Vac / 100 - 120VA

My question was: Do you think the transistors will survive when the input power is +/-12VDC
and the output is a 4 Ohm transformer load?

GL.
   
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From GL
Quote
My question was: Do you think the transistors will survive when the input power is +/-12VDC
and the output is a 4 Ohm transformer load?

I think the transistors will be well below the SOA (safe operating area) considering the low voltage.

I would put a fuse and a small 12 volt automobile lamp on the output (in series with the transformer) for initial testing, then later substitute a 1 ohm CSR so you can actually measure the current supplied to the transformer. Lamps can be dandy limiters if operated within their ratings.

If only a resistive load on your transformer, you will be safe. If you are going to be playing with resonance on the secondary, all bets are off as you may get severe resonance rise of voltage on the primary. In this case MOV's or tranzorbs may also be helpful to protect your amp from destruction.

Make sure your transformer can handle (has enough inductance) for the lowest frequency and power you intend to pass through it without saturating, as this could cause current spiking. The 12 V lamp will save the day here.

This amp is a nice basic design, and has built in current limiting (T4, T5), but this may not work under all conditions.

Be sure to conservatively fuse your batteries.

Prevention is prudent. Best of luck.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
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He is using series-opposing WW. As most of us do that don't have expensive current probes.

The problem is that he refuses to accept that in doing so, the polarity of the computed power will be backwards, and he is relying heavily on that negative number for his claim that not only is he powering a load, but he is returning excess energy back to the grid.  C.C

Oh my!

Sad. If he understood how the scope worked he would accept how you describe it.
As far as expensive probes go... Sure, I have access to some but almost never use them. Using floating measurement methods are easy and accurate enough for all but the most demanding tests.
   
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From GL
I think the transistors will be well below the SOA (safe operating area) considering the low voltage.

I would put a fuse and a small 12 volt automobile lamp on the output (in series with the transformer) for initial testing, then later substitute a 1 ohm CSR so you can actually measure the current supplied to the transformer. Lamps can be dandy limiters if operated within their ratings.

If only a resistive load on your transformer, you will be safe. If you are going to be playing with resonance on the secondary, all bets are off as you may get severe resonance rise of voltage on the primary. In this case MOV's or tranzorbs may also be helpful to protect your amp from destruction.

This amp has built in current limiting, but this may not work under all conditions.

Prevention is prudent. Best of luck.

ION,

Thank you for taking time to answer my questions.
The amplifier does not cost much and if I blow it up then no major loss. LOL

I probably has to wind the output transformer myself to get a 4 Ohm load.

GL.
   
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ION,

Thank you for taking time to answer my questions.
The amplifier does not cost much and if I blow it up then no major loss. LOL

I probably has to wind the output transformer myself to get a 4 Ohm load.

GL.


If operating at 60 Hz,y ou can probably use a standard off the shelf 120 Volt to 10 Volts (or thereabouts) transformer. No need to wind.

If I blow up an amp, I force myself to repair it (which I don't like doing) but it makes me more careful the next time around, my self imposed punishment. I would have to throw it away, which I would never do, thus it would lay around the shop haunting me.
This is why I am scrupulously careful. :D


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
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If operating at 60 Hz,y ou can probably use a standard off the shelf 120 Volt to 10 Volts (or thereabouts) transformer. No need to wind.

If I blow up an amp, I force myself to repair it (which I don't like doing) but it makes me more careful the next time around, my self imposed punishment. I would have to throw it away, which I would never do, thus it would lay around the shop haunting me.
This is why I am scrupulously careful. :D

ION,

I know what you are talking about. :-)

I just repaired four Shark-12 amplifiers for a friend. The bass output stage of those amplifiers
just blow up and burn if the amplifier gets to hot. So I just removed the burnt bass amplifier
parts and did use the posted amplifier (Wellman) instead. There was just enough room
inside the active speaker for the inserted amplifier. Got a mail from my friend that he had tested
the Shark-12 active speakers on a "gig" for 6 hours and was very pleased with the result.

I also did repaired a bass booster active speaker. Inside the speaker there was two transistors
(one NPN and one PNP) used to make a +/- 15 Volt power supply. Both transistors was soldered
into a PCB and there was mounted a small aluminum plate onto the transistors for cooling. The
aluminum plate was not secured to anything. So over the years the bass vibrations did vibrate
the aluminum plate back and forth until one of the transistors legs broke. I have never seen
that before. :-)

GL.
   

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It's not as complicated as it may seem...
In response to a reply by Farmhand from across the pond:


The scope can and will provide an accurate measurement regardless of the waveform, PROVIDED that there is little to no parasitic reactance between the measurement points, in particular the CSR.

But most importantly:

Correct interpretation of the results comes down to a basic understanding of electrical/electronics theory. More often than not, the simplest most basic concepts elude not only the uninitiated, but even the so-called trained individuals. And we are witnessing this phenomenon here.

Claims are being made based on (to quote from my good pal TK) "numbers in boxes", as provided by the scope itself. "A negative MEAN power must mean that a net power is returning to the grid" is the song being sung by so many uninitiated here, and the truth is that this is not only incorrect, but they have no idea why they are even singing this song. None of them truly understand the concepts of power measurement and how the measurement of sources and loads differ, not to mention adequately understanding how to correctly use the measurement features in their oscilloscope.

The claims and "tutorial" demonstrations being made here are, sadly, full of follies. Stay tuned for some real tutorials.  ;)
   
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